Answer: D let the length of the rectangular plot be x and breadth be y m. Given that, perimeter = 34. 2(x+y) = 34 => x+y = 17..........(i) Given that area = 60 xy = 60................(ii) (x-y)2 = (x+y)2 - 4xy = (17)2 - 4(60) = 49 => x- y = 7.............(iii) From equation (ii) and (iii), we get y = 5 m and x = 12 i.e shorter side be of 5 m.
Q. No. 68:
A lawn is in the form of an isosceles triangle. The cost of turfing it came to Rs 1200 at Rs 4 per m2 . If the base of the lawn is 40 m long, then the length of each side is
Answer: A Area of triangle = 1200/4 = 300 m2 Also, area of isosceles triangle = b/4 * (4a2 - b2 )1/2 Here, b = 40 m i.e 40/4 * (4a2 -1600)1/2 = 300 => 4a2 -1600 = 900 => a2= 625 => a = 25. Length of each side be 25 m.
Q. No. 69:
In the figure given below, the radius of the inscribed circle with the centre O is 10 cm. If the triangle ABC is equilateral and its side is a, find the area of triangle ABC (in cm2 ). Please note that AB, BC and CA are tangential to the circle.
Answer: A Given the radius of triangle ABC i.e OD = 10 cm.
Since, Triangle ABC is an equilateral, then AD = 3(OD) AD = 3* 10 = 30 cm. Now, in right angled triangle ABD a2 - a2 /4 = (30)2 => a = 20√3 Area of an equilateral triangle = √3/4 * a2 Area = 300√3 = 519.6 cm2
Q. No. 70:
Each side of a regular hexagon measures a cm. By joining the mid points of each side, another hexagon is formed inside it. The ratio of the areas of the outer and inner hexagons would be
Answer: B Let the length of circuit be l. 2(l+w) = p........(i) And, lw = k => l = k/w.......(ii) From both the equations, 2(k/w + w) = p 2(k + w2 ) = pw 2w2 + 2k - pw = 0.
Q. No. 72:
In the figure above, ABCD is a regular rectangle and its length is twice its width. If the area of triangle EAB is 144 cm2. What is the length of the rectangle?
